Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation cos(x)=-(1/2)
- Equation (x^2-81)^2+(x^2+5x-36)^2=0
-
Equation sin(x)=-√3/2
-
Equation 3t^2-4t+3=0
- Express {x} in terms of y where:
- 14*x-16*y=-3
- 17*x-20*y=6
- 10*x-19*y=3
- -2*x-1*y=-8
-
Identical expressions
- three t^ two -4t+3= zero
- 3t squared minus 4t plus 3 equally 0
- three t to the power of two minus 4t plus 3 equally zero
- 3t2-4t+3=0
- 3t²-4t+3=0
- 3t to the power of 2-4t+3=0
- 3t^2-4t+3=O
-
Similar expressions
- 3t^2+4t+3=0
- 3t^2-4t-3=0
3t^2-4t+3=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
$$a\ t^2 + b\ t + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$t_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$t_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 3$$
$$b = -4$$
$$c = 3$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 3 \cdot 4 \cdot 3 + \left(-4\right)^{2} = -20$$
Because D<0, then the equation
has no real roots,
but complex roots is exists.
$$t_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$t_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$t_{1} = \frac{2}{3} + \frac{\sqrt{5} i}{3}$$
Simplify
$$t_{2} = \frac{2}{3} - \frac{\sqrt{5} i}{3}$$
Simplify
$$a\ t^2 + b\ t + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$t_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$t_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 3$$
$$b = -4$$
$$c = 3$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 3 \cdot 4 \cdot 3 + \left(-4\right)^{2} = -20$$
Because D<0, then the equation
has no real roots,
but complex roots is exists.
$$t_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$t_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$t_{1} = \frac{2}{3} + \frac{\sqrt{5} i}{3}$$
Simplify
$$t_{2} = \frac{2}{3} - \frac{\sqrt{5} i}{3}$$
Simplify
Vieta's Theorem
rewrite the equation
$$3 t^{2} - 4 t + 3 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{4 t}{3} + 1 = 0$$
$$p t + t^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{4}{3}$$
$$q = \frac{c}{a}$$
$$q = 1$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{4}{3}$$
$$t_{1} t_{2} = 1$$
$$3 t^{2} - 4 t + 3 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{4 t}{3} + 1 = 0$$
$$p t + t^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{4}{3}$$
$$q = \frac{c}{a}$$
$$q = 1$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{4}{3}$$
$$t_{1} t_{2} = 1$$
Rapid solution
[src]
___
2 I*\/ 5
t_1 = - - -------
3 3
$$t_{1} = \frac{2}{3} - \frac{\sqrt{5} i}{3}$$
___
2 I*\/ 5
t_2 = - + -------
3 3
$$t_{2} = \frac{2}{3} + \frac{\sqrt{5} i}{3}$$
Sum and product of roots
[src]
sum
___ ___ 2 I*\/ 5 2 I*\/ 5 - - ------- + - + ------- 3 3 3 3
$$\left(\frac{2}{3} - \frac{\sqrt{5} i}{3}\right) + \left(\frac{2}{3} + \frac{\sqrt{5} i}{3}\right)$$
=
4/3
$$\frac{4}{3}$$
product
___ ___ 2 I*\/ 5 2 I*\/ 5 - - ------- * - + ------- 3 3 3 3
$$\left(\frac{2}{3} - \frac{\sqrt{5} i}{3}\right) * \left(\frac{2}{3} + \frac{\sqrt{5} i}{3}\right)$$
=
1
$$1$$
Numerical answer
[src]
t1 = 0.666666666666667 - 0.74535599249993*i
t2 = 0.666666666666667 + 0.74535599249993*i
t2 = 0.666666666666667 + 0.74535599249993*i
The graph